A = 3x² - 7x + 8 = 3(x² - 7/3 + 49/36) + 47/12 = 3(x - 7/6)² + 47/12

Ta luôn có: 3(x - 7/6)² \(\ge\)0 \(\forall\)x

=> 3(x - 7/6)² + 47/12 \(\ge\)47/12 \(\forall\)x

Dấu "=" xảy ra khi: x - 7/6 = 0 <=> x = 7/6

Vậy Min của A = 47/12 tại x = 7/6

A= X²+5X+25/4-37/4 =(X+5/2)²-37/4 >= -37/4

A_min=-37/4

Đạt được khi : X=-5/2

B=-X²+7X+1=-(X²-7X-1)=-(X²+7X+49/4-53/4)=-(X+7/2)²+53/4<=53/4

B_Max=53/4

Đạt được khi:X=-7/2

C=2x²+6x=2x²+6x+9/4-9/4=2(x²+3x+9/4)-9/4=2(x+3/2)²-9/4>=-9/4

C_Min=-9/4

Đạt được khi:x=-3/2

Đặt \(A=7x^2+5x+3\)

\(A=\left(7x^2+5x+\frac{25}{28}\right)+\frac{59}{28}\)

\(A=7\left(x^2+\frac{5}{7}x+\frac{25}{196}\right)+\frac{59}{28}\)

\(A=7\left(x+\frac{5}{14}\right)^2+\frac{59}{28}\ge\frac{59}{28}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(7\left(x+\frac{5}{14}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-5}{14}\)

Vậy GTNN của \(A\) là \(\frac{59}{28}\) khi \(x=\frac{-5}{14}\)

Đặt \(B=-3x^2-3x+5\)

\(B=\left(-3x^2-3x-\frac{3}{4}\right)+\frac{23}{4}\)

\(B=-3\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}\)

\(B=-3\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\le\frac{23}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-3\left(x+\frac{1}{2}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-1}{2}\)

Vậy GTLN của \(B\) là \(\frac{23}{4}\) khi \(x=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

Ta có:

\(7x^2+5x+3=7\left(x^2+\frac{5}{7}x+\frac{3}{7}\right)\)

\(=7\left(x^2+\frac{5}{7}x+\frac{25}{196}+\frac{59}{196}\right)\)

\(=7\left(x+\frac{5}{14}\right)^2+\frac{59}{28}\ge\frac{59}{28}\)

\(-3x^2-3x+5=-3\left(x^2+x-5\right)\)

\(=-3\left(x^2+x+\frac{1}{4}-\frac{21}{4}\right)=-3\left(x+\frac{1}{2}\right)^2+\frac{63}{4}\le\frac{63}{4}\)

a) Ta có: \(7x^2+5x+3=7\left(x^2+\frac{5}{7}x\right)+3\)

\(=7\left(x^2+2.\frac{5}{14}x+\frac{25}{196}\right)+3=7\left(x+\frac{5}{14}\right)^2+\frac{59}{28}\ge\frac{59}{28}\)

Dấu "=" xảy ra \(\Leftrightarrow7\left(x+\frac{5}{14}\right)^2=0\Leftrightarrow x=\frac{-5}{14}\)

Vậy giá trị nhỏ nhất của biểu thức là ....

b) \(-3x^2-3x+5=-\left(3x^2+3x-\frac{3}{4}\right)+\frac{23}{4}\)

\(=-3\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}\)

\(=-3\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\le\frac{23}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy GTLN của biểu thức trên là ...

a: \(f\left(x\right)=2x^2-7x+9\)

=>\(f'\left(x\right)=2\cdot2x-7=4x-7\)

Đặt f'(x)=0

=>\(4x-7=0\)

=>\(x=\dfrac{7}{4}\)

\(f\left(\dfrac{7}{4}\right)=2\cdot\left(\dfrac{7}{4}\right)^2-7\cdot\dfrac{7}{4}+9=\dfrac{23}{8}\)

\(f\left(-1\right)=2\left(-1\right)^2-7\cdot\left(-1\right)+9=18\)

\(f\left(4\right)=2\cdot4^2-7\cdot4+9=13\)

Vì \(f\left(\dfrac{7}{4}\right)< f\left(4\right)< f\left(-1\right)\)

nên \(f\left(x\right)_{max\left[-1;4\right]}=18;f\left(x\right)_{min\left[-1;4\right]}=\dfrac{23}{8}\)

b: \(f\left(x\right)=x^2+5x+3\)

=>\(f'\left(x\right)=2x+5\)

f'(x)=0

=>2x+5=0

=>2x=-5

=>\(x=-\dfrac{5}{2}\)

\(f\left(-\dfrac{5}{2}\right)=\left(-\dfrac{5}{2}\right)^2+5\cdot\dfrac{-5}{2}+3=\dfrac{25}{4}-\dfrac{25}{2}+3=-\dfrac{13}{4}\)

\(f\left(2\right)=2^2+5\cdot2+3=4+10+3=17\)

\(f\left(6\right)=6^2+5\cdot6+3=69\)

Vậy: \(f\left(x\right)_{max\left[2;6\right]}=69;f\left(x\right)_{min\left[2;6\right]}=-\dfrac{13}{4}\)

a) \(A=25x^2-10x+9\)

\(A=\left(5x\right)^2-2\cdot5x\cdot1+1^2+9\)

\(A=\left(5x-1\right)^2+9\ge9\)

Dấu "=" xảy ra \(\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)